1.

100 ml of PH_(3) on heating forms P and H_(2). The volume change in the reaction is

Answer»

an INCREASE of 50 ML
an increase of 100 ml
an increase of 150 ml
a decrease of 50 ml

Solution :`UNDERSET("2 mL")(2PH_(3))(g)rarr2P(s)+underset("3 mL")(3H_(2)(g))`
2 mL `PH_(3)` on decomposition give 3 mL of `H_(2)`, i.e., increase = 1 mL
`therefore "100 mL PH"_(3)" will result in increase = 50 mL"`


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